Series: determining convergence of series problem
$begingroup$
I am working on the following assignment about series:
$$ a_n , , is ,, a ,, convergent ,, progression ,, and \
sum_1^infty (a_n+1) = lim_{xtoinfty} {x^2-x+1over x^2-cosx}$$ Determine if the sum converges and $$lim_{xtoinfty} a_n$$
What I have done so far:
I know that for a sum to converge, its associated progression must have 0 as its limit at infinity (the neccessary condition).
I also know that, if the limit of the sum at infinity exists, then the sum converges and that limit is the value of the sum. My first step was finding the RHS limit:
$$lim_{xtoinfty} {x^2-x+1over x^2-cosx} = 1 $$
Now, this means -to me- that the sum is convergent, and it converges to 1 . However, I find the following problem. I also know that summations are "linear", I mean:
$$sum_1^infty a_n+1 = sum_1^infty a_n+ sum_1^infty 1 $$
I know the first one is convergent, but the second one clearly is not! Its limit at infinity is infinity; and anything + infinity = infinity = divergent. I cannot decide where was my mistake. Could someone help me to figure it out?
Lots of thanks in advance!
sequences-and-series
asked Dec 27 '18 at 1:49
ArtemArtem
425
$endgroup$
|
show 3 more comments
$begingroup$
I am working on the following assignment about series:
$$ a_n , , is ,, a ,, convergent ,, progression ,, and \
sum_1^infty (a_n+1) = lim_{xtoinfty} {x^2-x+1over x^2-cosx}$$ Determine if the sum converges and $$lim_{xtoinfty} a_n$$
What I have done so far:
I know that for a sum to converge, its associated progression must have 0 as its limit at infinity (the neccessary condition).
I also know that, if the limit of the sum at infinity exists, then the sum converges and that limit is the value of the sum. My first step was finding the RHS limit:
$$lim_{xtoinfty} {x^2-x+1over x^2-cosx} = 1 $$
Now, this means -to me- that the sum is convergent, and it converges to 1 . However, I find the following problem. I also know that summations are "linear", I mean:
$$sum_1^infty a_n+1 = sum_1^infty a_n+ sum_1^infty 1 $$
I know the first one is convergent, but the second one clearly is not! Its limit at infinity is infinity; and anything + infinity = infinity = divergent. I cannot decide where was my mistake. Could someone help me to figure it out?
Lots of thanks in advance!
sequences-and-series
asked Dec 27 '18 at 1:49
ArtemArtem
425
$endgroup$
$begingroup$
See my answer below.
$endgroup$
– John Omielan
Dec 27 '18 at 1:54
1
$begingroup$
Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 1:59
$begingroup$
$$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:00
$begingroup$
@rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
$endgroup$
– Artem
Dec 27 '18 at 2:21
1
$begingroup$
You knew $$sum_1^infty (a_n+1 )= 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:30
|
show 3 more comments
$begingroup$
I am working on the following assignment about series:
$$ a_n , , is ,, a ,, convergent ,, progression ,, and \
sum_1^infty (a_n+1) = lim_{xtoinfty} {x^2-x+1over x^2-cosx}$$ Determine if the sum converges and $$lim_{xtoinfty} a_n$$
What I have done so far:
I know that for a sum to converge, its associated progression must have 0 as its limit at infinity (the neccessary condition).
I also know that, if the limit of the sum at infinity exists, then the sum converges and that limit is the value of the sum. My first step was finding the RHS limit:
$$lim_{xtoinfty} {x^2-x+1over x^2-cosx} = 1 $$
Now, this means -to me- that the sum is convergent, and it converges to 1 . However, I find the following problem. I also know that summations are "linear", I mean:
$$sum_1^infty a_n+1 = sum_1^infty a_n+ sum_1^infty 1 $$
I know the first one is convergent, but the second one clearly is not! Its limit at infinity is infinity; and anything + infinity = infinity = divergent. I cannot decide where was my mistake. Could someone help me to figure it out?
Lots of thanks in advance!
sequences-and-series
asked Dec 27 '18 at 1:49
ArtemArtem
425
$endgroup$
I am working on the following assignment about series:
$$ a_n , , is ,, a ,, convergent ,, progression ,, and \
sum_1^infty (a_n+1) = lim_{xtoinfty} {x^2-x+1over x^2-cosx}$$ Determine if the sum converges and $$lim_{xtoinfty} a_n$$
What I have done so far:
I know that for a sum to converge, its associated progression must have 0 as its limit at infinity (the neccessary condition).
I also know that, if the limit of the sum at infinity exists, then the sum converges and that limit is the value of the sum. My first step was finding the RHS limit:
$$lim_{xtoinfty} {x^2-x+1over x^2-cosx} = 1 $$
Now, this means -to me- that the sum is convergent, and it converges to 1 . However, I find the following problem. I also know that summations are "linear", I mean:
$$sum_1^infty a_n+1 = sum_1^infty a_n+ sum_1^infty 1 $$
I know the first one is convergent, but the second one clearly is not! Its limit at infinity is infinity; and anything + infinity = infinity = divergent. I cannot decide where was my mistake. Could someone help me to figure it out?
Lots of thanks in advance!
sequences-and-series
sequences-and-series
asked Dec 27 '18 at 1:49
ArtemArtem
425
asked Dec 27 '18 at 1:49
ArtemArtem
425
edited Dec 27 '18 at 2:23
Artem
asked Dec 27 '18 at 1:49
ArtemArtem
425
asked Dec 27 '18 at 1:49
ArtemArtem
425
asked Dec 27 '18 at 1:49
ArtemArtem
425
425
$begingroup$
See my answer below.
$endgroup$
– John Omielan
Dec 27 '18 at 1:54
1
$begingroup$
Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 1:59
$begingroup$
$$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:00
$begingroup$
@rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
$endgroup$
– Artem
Dec 27 '18 at 2:21
1
$begingroup$
You knew $$sum_1^infty (a_n+1 )= 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:30
|
show 3 more comments
$begingroup$
See my answer below.
$endgroup$
– John Omielan
Dec 27 '18 at 1:54
1
$begingroup$
Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 1:59
$begingroup$
$$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:00
$begingroup$
@rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
$endgroup$
– Artem
Dec 27 '18 at 2:21
1
$begingroup$
You knew $$sum_1^infty (a_n+1 )= 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:30
$begingroup$
See my answer below.
$endgroup$
– John Omielan
Dec 27 '18 at 1:54
$begingroup$
See my answer below.
$endgroup$
– John Omielan
Dec 27 '18 at 1:54
1
1
$begingroup$
Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 1:59
$begingroup$
Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 1:59
$begingroup$
$$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:00
$begingroup$
$$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:00
$begingroup$
@rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
$endgroup$
– Artem
Dec 27 '18 at 2:21
$begingroup$
@rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
$endgroup$
– Artem
Dec 27 '18 at 2:21
1
1
$begingroup$
You knew $$sum_1^infty (a_n+1 )= 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:30
$begingroup$
You knew $$sum_1^infty (a_n+1 )= 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:30
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I believe the expression
$$sum_{1}^{infty} a_n + 1$$
means
$$left(sum_{1}^{infty} a_n right) + 1$$
If it meant as you suggest, the expression would instead most likely be written as
$$sum_{1}^{infty} left( a_n + 1 right)$$
This would avoid potential ambiguity.
answered Dec 27 '18 at 1:58
John OmielanJohn Omielan
3,9251215
$endgroup$
$begingroup$
Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
$endgroup$
– Artem
Dec 27 '18 at 2:10
$begingroup$
@Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
$endgroup$
– John Omielan
Dec 27 '18 at 2:13
$begingroup$
@Artem Well, the sum converges if it is equal to 1...
$endgroup$
– Osvaldo Paniccia
Dec 27 '18 at 2:27
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
I believe the expression
$$sum_{1}^{infty} a_n + 1$$
means
$$left(sum_{1}^{infty} a_n right) + 1$$
If it meant as you suggest, the expression would instead most likely be written as
$$sum_{1}^{infty} left( a_n + 1 right)$$
This would avoid potential ambiguity.
answered Dec 27 '18 at 1:58
John OmielanJohn Omielan
3,9251215
$endgroup$
$begingroup$
Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
$endgroup$
– Artem
Dec 27 '18 at 2:10
$begingroup$
@Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
$endgroup$
– John Omielan
Dec 27 '18 at 2:13
$begingroup$
@Artem Well, the sum converges if it is equal to 1...
$endgroup$
– Osvaldo Paniccia
Dec 27 '18 at 2:27
add a comment |
$begingroup$
I believe the expression
$$sum_{1}^{infty} a_n + 1$$
means
$$left(sum_{1}^{infty} a_n right) + 1$$
If it meant as you suggest, the expression would instead most likely be written as
$$sum_{1}^{infty} left( a_n + 1 right)$$
This would avoid potential ambiguity.
answered Dec 27 '18 at 1:58
John OmielanJohn Omielan
3,9251215
$endgroup$
$begingroup$
Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
$endgroup$
– Artem
Dec 27 '18 at 2:10
$begingroup$
@Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
$endgroup$
– John Omielan
Dec 27 '18 at 2:13
$begingroup$
@Artem Well, the sum converges if it is equal to 1...
$endgroup$
– Osvaldo Paniccia
Dec 27 '18 at 2:27
add a comment |
$begingroup$
I believe the expression
$$sum_{1}^{infty} a_n + 1$$
means
$$left(sum_{1}^{infty} a_n right) + 1$$
If it meant as you suggest, the expression would instead most likely be written as
$$sum_{1}^{infty} left( a_n + 1 right)$$
This would avoid potential ambiguity.
answered Dec 27 '18 at 1:58
John OmielanJohn Omielan
3,9251215
$endgroup$
I believe the expression
$$sum_{1}^{infty} a_n + 1$$
means
$$left(sum_{1}^{infty} a_n right) + 1$$
If it meant as you suggest, the expression would instead most likely be written as
$$sum_{1}^{infty} left( a_n + 1 right)$$
This would avoid potential ambiguity.
answered Dec 27 '18 at 1:58
John OmielanJohn Omielan
3,9251215
edited Dec 27 '18 at 2:05
answered Dec 27 '18 at 1:58
John OmielanJohn Omielan
3,9251215
answered Dec 27 '18 at 1:58
John OmielanJohn Omielan
3,9251215
answered Dec 27 '18 at 1:58
John OmielanJohn Omielan
3,9251215
3,9251215
$begingroup$
Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
$endgroup$
– Artem
Dec 27 '18 at 2:10
$begingroup$
@Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
$endgroup$
– John Omielan
Dec 27 '18 at 2:13
$begingroup$
@Artem Well, the sum converges if it is equal to 1...
$endgroup$
– Osvaldo Paniccia
Dec 27 '18 at 2:27
add a comment |
$begingroup$
Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
$endgroup$
– Artem
Dec 27 '18 at 2:10
$begingroup$
@Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
$endgroup$
– John Omielan
Dec 27 '18 at 2:13
$begingroup$
@Artem Well, the sum converges if it is equal to 1...
$endgroup$
– Osvaldo Paniccia
Dec 27 '18 at 2:27
$begingroup$
Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
$endgroup$
– Artem
Dec 27 '18 at 2:10
$begingroup$
Hey, thanks for the answer. Nope, I asked the lecturer and he said it must be considered with brackets!
$endgroup$
– Artem
Dec 27 '18 at 2:10
$begingroup$
@Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
$endgroup$
– John Omielan
Dec 27 '18 at 2:13
$begingroup$
@Artem You are welcome. You may wish to suggest to your lecturer that he/she uses brackets to avoid this potential ambiguity.
$endgroup$
– John Omielan
Dec 27 '18 at 2:13
$begingroup$
@Artem Well, the sum converges if it is equal to 1...
$endgroup$
– Osvaldo Paniccia
Dec 27 '18 at 2:27
$begingroup$
@Artem Well, the sum converges if it is equal to 1...
$endgroup$
– Osvaldo Paniccia
Dec 27 '18 at 2:27
add a comment |
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$begingroup$
See my answer below.
$endgroup$
– John Omielan
Dec 27 '18 at 1:54
1
$begingroup$
Perhaps the progression ${a_n}$ converges to $-1$ : $$limlimits_{ntoinfty}a_n=-1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 1:59
$begingroup$
$$sum_1^infty( -1+1)~~~ ?=~~~ sum_1^infty -1+ sum_1^infty 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:00
$begingroup$
@rsadhvika I initially thought the same thing but, wouldn't that contradict the first limit found?
$endgroup$
– Artem
Dec 27 '18 at 2:21
1
$begingroup$
You knew $$sum_1^infty (a_n+1 )= 1$$
$endgroup$
– rsadhvika
Dec 27 '18 at 2:30